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insens350 [35]
2 years ago
15

The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of which fun

ction(s)? tangent only cosine only sine and tangent cosine and sine
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
4 0

The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

<h3>What is vertical motion of a projecile?</h3>

The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;

Vy = Vsinθ

<h3>What is horizontal motion of a projecile?</h3>

The horizontal velocity of a projectile is given by the following formula;

Vx = Vcosθ

<h3>Direction of the motion</h3>

The direction of the motion is calculated as follows;

tanθ  = Vy/Vx

Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

Learn more about vertical motion here: brainly.com/question/24216590

#SPJ4

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Answer:

The force of friction acting on block B is approximately 26.7N.  Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.  

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation  is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

a_B = a_A-0.5 \frac{m}{s^2}

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

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